Find Domain of Log Function
Need help determining the Domain of $$\log_2\left(2x+3\right)=\log_2\left(3x+4\right)-\log_2\left(3x+2\right)$$
Do not know if I should simplify first to get the domain, or to get the domain from the set logs that are in place?
$\endgroup$ 42 Answers
$\begingroup$This is an equation, and you can solve it. First of all we need to understand the single domains. That is easy. I will use the writing $\lg$ for the log base $2$.
$$\lg(2x+3)$$
This function exists iff $2x+3 >0$ hence $\boxed{x > -\frac{3}{2}}$.
$$\lg(3x+4)$$
This function exists if $3x+4 >0$ hence $\boxed{x > -\frac{4}{3}}$
$$\lg(3x+2)$$
This function exists if $3x+2 >0$ hence $\boxed{x >-\frac{2}{3}}$
Finally you can write th equation
$$\lg(2x+3) = \lg\left[(3x+4)/(3x+2)\right]$$
Where I used the well known property of the sum of log.
So now you can exponentiate in base $2$:
$$2x+3 = \frac{3x+4}{3x+2}$$
$$(2x+3)(3x+2) = 3x+4$$
$$6x^2 + 13x + 6 = 3x + 4$$
$$6x^2 + 10x + 2 = 0$$
Now you can solve it as a normal second degree equation
$$x = \frac{-10\pm\sqrt{100 - 48}}{12}$$
And check for the solutions, that don't have to be in contrast with the previous assumptions (necessarily).
$\endgroup$ $\begingroup$I see an equation, not (just) a function?
In any case: $\log_a x$ (for a positive base $a$, different from $1$) is only defined for $x>0$. So for any logarithm appearing in your equation, the argument of the logarithm has to be strictly positive.
E.g. for $\log_2(3x+2)$ you need $3x+2>0 \Leftrightarrow x > -\tfrac{2}{3}$.
For all logarithms to be defined, you'll want the intersection of the 'domains' (sets where the different logarithms are defined), before simplifying (see below).
Notice that $$\log x + \log x$$ is defined only for $x>0$ while (use $\log a + \log b = \log (ab)$) $$\log \left( x^2 \right)$$ is defined for all $x\ne0$.
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